mHCl= 10,95/36,5=0,3(mol)

Vì Cu ko phản ứng vs HCl nên suy ra

Mg+2HCl--->MgCl2+H2

0,15<-- 0,3 --> 0,15.(Mol)

mMg=0,15x24=3,6(g)

%Mg=3,6.100%/10=36%

%Cu=100%-36%=64%

b, mdd=10+300=310 g

mMgCl2=0,15x24=3,6 g

C%MgCl2 =3,6/310x100%≈1,16%

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2               1         1

       0,05    0,1           0,05      0,05 

    \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

       1           2              1            1

      0,2       0,4            0,2

a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1,2\left(g\right)\)

\(m_{MgO}=9,2-1,2=8\left(g\right)\)

⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

b) Có : \(m_{MgO}=8\left(g\right)\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)

\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)

\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)⁰/₀

 Chúc bạn học tốt

a)\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05    0,1           0,05        0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:     0,2         0,4         0,2

\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)

\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)

b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c,m_{dd sau pứ} = 9,2+125-0,05.2 = 134,1 (g)

 \(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)

\(Mg + 2HCl \rightarrow MgCl_2 + H_2\)     (1)

\(MgO + 2HCl \rightarrow MgCl_2 + H_2O\)      (2)

Khí thu được là H₂

\(n_{H_2}= \dfrac{1,12}{22,4}=0,05 mol\)

Theo PTHH (1):

\(n_{Mg}= n_{H_2}= 0,05 mol\)

\(\Rightarrow m_{Mg}= 0,05 . 24= 1,2 g\)

\(\Rightarrow m_{MgO}= 9,2 - 1,2= 8g\)

C%_\(Mg\)= \(\dfrac{1,2}{9,2} .100\)%=13,04%

C%_\(MgO\)= 100% - 13,04%=86,96%

b)

\(n_{MgO}= \dfrac{8}{40}=0,2 mol\)

Theo PTHH (1) và (2):

\(n_{HCl(1)}= 2n_{Mg}= 0,1 mol\)

\(n_{HCl(2)}= 2n_{MgO}= 0,4 mol\)

Suy ra: \(n_{HCl}= n_{HCl(1)} + n_{HCl(2)}\)

                     \(= 0,1 + 0,4= 0,5 mol\)

\(\Rightarrow m_{HCl}= 0,5 . 36,5= 18,25g\)

\(\Rightarrow m_{dd HCl} = \dfrac{18,25 . 100%}{14,6%}=125 g\)

_c)

\(\)Dung dịch sau pư: MgCl₂

Theo PTHH:

\(n_{MgCl_2}= \dfrac{1}{2} n_{HCl}= 0,25 mol\)

\(\Rightarrow m_{MgCl_2}= 0,25 . 95=23,75g\)

\(m_{dd sau pư} = m_{Mg} + m_{MgO} + m_{dd HCl}- m_{H_2}\)

               \(= 9,2 + 125 + 2 . 0,05\)

             \(=134,1 g\)

C%_{\(MgCl_2\)}=\(\dfrac{23,75}{134,1}. 100\)%=17,71%

- Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow27a+24b=10,2\left(1\right)\)

Khí thu được sau p/ứ là khí H₂: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

2                                         3       (mol)

a                                        3/2 a   (mol)

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

1                                        1   (mol)

b                                         b   (mol)

Từ hai PTHH trên ta có: \(\dfrac{3}{2}a+b=0,5\left(2\right)\)

\(\left(1\right),\left(2\right)\) ta có hệ: \(\left\{{}\begin{matrix}27a+24b=10,2\\\dfrac{3}{2}a+b=0,5\end{matrix}\right.\)

Giải ra ta có \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

a) \(\%Al=\dfrac{m_{Al}}{m_{hh}}.100\%=\dfrac{0,2.27}{10,2}.100\%\approx52,94\%\)

\(\%Mg=100\%-\%Al=100\%-52,94=47,06\%\)

b)

 \(3H_2+Fe_2O_3\rightarrow^{t^0}2Fe+3H_2O\)

3               1              2  (mol)

0,5            1/6         1/3  (mol)

\(m_{Fe}=\dfrac{1}{3}.56=\dfrac{56}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(pứ\right)}=\dfrac{1}{6}.160=\dfrac{80}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(dư\right)}=60-m_{Fe}=60-\dfrac{56}{3}=\dfrac{124}{3}\left(g\right)\)

\(a=\dfrac{124}{3}+\dfrac{80}{3}=68\left(g\right)\)

a)

PTHH: 2A + 2nHCl --> 2ACl_n + nH₂

            2B + 2mHCl --> 2BCl_m + mH₂

Gọi số mol H₂ là a (mol)

=> n_HCl = 2a (mol)

Theo ĐLBTKL: m_{kim loại} + m_HCl = m_muối + m_H2

=> 8,9 + 36,5.2a = 23,1 + 2a

=> a = 0,2 (mol)

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) 

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

=> \(n_{H_2\left(tăng\right)}=0,25-0,2=0,05\left(mol\right)\)

PTHH: 2B + 2mHCl --> 2BCl_m + mH₂

           \(\dfrac{0,1}{m}\)<------------\(\dfrac{0,1}{m}\)<---0,05

Khối lượng rắn sau pư tăng lên do có thêm BCl_m sinh ra

=> \(m_{BCl_m}=\dfrac{0,1}{m}\left(M_B+35,5m\right)=27,85-23,1=4,75\left(g\right)\)

=> M_B = 12m (g/mol)

Xét m = 2 thỏa mãn => M_B = 24 (g/mol) => B là Mg

\(n_{Mg\left(thêm\right)}=\dfrac{0,1}{m}=\dfrac{0,1}{2}=0,05\left(mol\right)\)

=> \(n_{Mg\left(bđ\right)}=0,1\left(mol\right)\)

=> \(m_A=8,9-0,1.24=6,5\left(g\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             0,1-------------------->0,1

             2A + 2nHCl --> 2ACl_n + nH₂

            \(\dfrac{0,2}{n}\)<-------------------0,1

=> \(M_A=\dfrac{6,5}{\dfrac{0,2}{n}}=32,5n\left(g/mol\right)\)

 Xét n = 2 thỏa mãn => M_A = 65 (g/mol)

=> A là Zn

 PTHH

Mg + 2HCl ----> MgCl2 + H2      (1)

MgO + 2HCl -----> MgCl2 + H2O (2)

a) Theo pt(1) n Mg = n H2 = \(\frac{1,12}{22,4}\) = 0,05 (mol)

==> m Mg = 0,005 . 24=1,2 (g)

%m Mg =  \(\frac{1,2}{3,2}\). 100%= 37,5%

%m MgO= 100% - 37,5%= 62,5%

b)m dd sau pư = 3,2 + 246,9 - 0,05 . 2=250 (g)

Theo pt(1)(2)  n MgCl2(1) = n Mg = 0,05 mol

                         n MgCl2 (2) = n MgO=\(\frac{3,2-1,2}{40}\)=0,05(mol)

==> tổng n MgCl2 = 0,1 (mol)  ---->m MgCl2 = 9,5 (g)

C%(MgCl2)= \(\frac{9,5}{250}\) .100% = 3,8%

Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O (1)

CuO + 2HCl → CuCl₂ + H₂O (2)

a) \(m_{CuO}=20\times20\%=4\left(g\right)\)

\(\Rightarrow m_{Fe_2O_3}=20-4=16\left(g\right)\)

b) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT1: \(n_{HCl}=6n_{Fe_2O_3}=6\times0,1=0,6\left(mol\right)\)

Theo PT2: \(n_{HCl}=2n_{CuO}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,1+0,6=0,7\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,7\times36,5=25,55\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{25,55}{5,475\%}=466,67\left(g\right)\)

c) Dung dịch sau phản ứng gồm: CuCl₂ và FeCl₃

Theo PT1: \(n_{FeCl_3}=2n_{Fe_2O_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=0,2\times162,5=32,5\left(g\right)\)

Theo PT2: \(n_{CuCl_2}=n_{CuO}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuCl_2}=0,05\times135=6,75\left(g\right)\)

\(\Sigma m_{dd}=20+466,67=486,67\left(g\right)\)
\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{486,67}\times100\%=6,68\%\)

\(C\%_{CuCl_2}=\dfrac{6,75}{486,67}\times100\%=1,39\%\)

Bài 3 :

\(a) n_{CuO} = a(mol) ; n_{Fe_2O_3} = b(mol)\\ \Rightarrow 80a + 160b = 36(1)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + H_2O\\ n_{Cu} = n_{CuO} = a(mol)\\ n_{Fe} = 2n_{Fe_2O_3} = 2b(mol)\\ \Rightarrow 64a = 4.2b.56(2)\\ (1)(2) \Rightarrow a = 0,35 ; b = 0,05\\ m_{CuO} = 0,35.80 = 28(gam)\\ m_{Fe_2O_3} = 0,05.160 = 8(gam)\\ b) n_{H_2} = a + 3b = 0,5(mol) \Rightarrow V_{H_2} = 0,5.22,4 = 11,2(lít)\)

\(c) Fe + 2HCl \to FeCl_2 + H_2\\ n_{HCl} = 2n_{Fe} = 0,1.2 = 0,2(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{10,95\%} = 66,67(gam)\)

Bài 4 :

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = n_{Zn} = \dfrac{1,95}{65} = 0,03(mol)\\ V_{H_2} = 0,03.22,4= 0,672(lít)\\ b) n_{HCl} =2 n_{H_2} = 0,06(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,06.36,5}{120}.100\% = 1,825\%\\ m_{dd\ sau\ pư} = 1,95 + 120 - 0,03.2 = 121,89(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,03.136}{121,89}.100\% = 3,35\%\)